Homework 2: Solutions GFD I Winter 2007
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1 Homework : Solutions GFD I Winter a. Part One The goal is to find the height that the free surface at the edge of a spinning beaker rises from its resting position. The first step of this process is to find an expression for the free surface height, η, as a function of radius, r, or distance from the center axis. In this interpretation of the problem, the z axis is parallel to the rotational vector Ω. The radial axis is perpendicular to the z axis, with r = 0 at the center of the beaker. True gravity points downward along the z axis. Here, z = 0 is defined as the flat resting surface of the fluid. The height η(r) is the distance between the fluid surface in a spinning beaker and z = 0. The depth of the fluid in a rest state is defined as z = -H. The radius of the beaker is R. One way to approach this problem is by using a force balance. The momentum equation in a rotating frame of reference is given by 1
2 du = 1 ρ p φ grav + Ω ri Ω u, where i denotes a unit vector along the r axis. In this case, since there is no velocity (in a rotating frame) and no acceleration, the momentum equation reduces to 0 = 1 ρ p gk + Ω ri, In the vertical direction, there is simply hydrostatic balance: 0 = 1 η ρ z g. Integrating this equation vertically from -H to η gives the pressure at the bottom, p = p( H) = ρg(η + H) p atm. In the horizontal, or radial, direction, the pressure gradient force balances the centrifugal acceleration: 0 = 1 p ρ r + Ω r. Substituting hydrostatic pressure p(-h), 0 = g η r + Ω r. Integrating in the radial direction gives an (incomplete) expression for surface height as a function of radius: η = Ω r g An alternative approach is to treat the free surface as having constant geopotential: + c
3 Part Two Φ = gz Ω r const. = gη Ω r η = Ω r g There are different methods to calculate the distance the free surface rises from its resting height, all involving volume conservation. The method presented here involves integration of circular shells. Since mass (and therefore volume in a constant density fluid) is conserved, volume displaced upward from the resting surface z = 0 must be removed from below the resting surface. + c R ( Ω r 0 = 0 g ( Ω R 4 = π 8g c = Ω R 4g ) + c πrdr ) + cr This means that difference between the resting surface and the height at the edge of the beaker is also Ω R 4g, one half of the total height difference between the center and edge. 1.b. In determining whether the bubbles will be closer together when they reach the surface, consider the bubbles at a time t = 0, just before they start to move. For a fluid parcel with no motion in a rotating reference frame, the full momentum equation is given by 0 = 1 ρ p gk + Ω ri. As seen in part a, the pressure gradients in the beaker are set up to balance true gravity and the centrifugal force. The bubbles will have much less density than the surrounding fluid, but the pressure exerted on them will be the same as any other fluid parcel located the same distance away from the center of the beaker. The first term in the above equation, the acceleration due to the pressure gradient force, will therefore increase in magnitude. Its direction stays the same. 3
4 Motion will be opposite of the direction of the combined force of gravity and centrifugal force. In the center, there is no centrifugal force, so the motion of the bubble will be entirely vertical. Away from the center, the motion start out up and inward. The motion of the bubbles is perpendicular to the geopotential surfaces on which they are located. Coriolis forces will only affect the bubble that is away from the center. Since the axis of rotation is vertical, the Coriolis force, Ω u, will not affect the strictly vertical motion of the center bubble. This is just like the North Pole, where vertical velocity does not induce a Coriolis force because it is parallel to the axis of rotation. The Coriolis force will deflect the horizontal motion of the outer bubble to the right. The magnitude of this force is proportional to the magnitude of the horizontal velocity of the bubble. So the bubble will feel no Coriolis force at time t = 0, then feel more Coriolis force as it accelerates inward and its horizontal velocity increases. The importance of Coriolis force will vary with the amount of rotation. If the bubbles are small, they will experience strong frictional forces. This will keep the velocity of the bubbles slow, so Coriolis forces should only play a very minor role compared to bouyancy forces. 1.c. In a two layer fluid, it is possible to calculate the interface height in the same way as the free surface height in part a. Define E as the displacement of the interface from its resting height. The height between the interface and the surface is h 1 and the height between the bottom and the interface is h. The free surface height will have the same shape as in part a and will not be affected by the fluid layer below. 4
5 There is hydrostatic balance in the vertical direction. bottom to the surface gives the pressure at the bottom, Integrating from the patm p η p = ρg z H p = ρ 1 gh 1 + ρ gh p atm At the bottom, since there is no velocity or acceleration in the rotating reference frame, the horizontal pressure gradient force balances the centrifugal force, as in part a. ρ Ω r = p r ρ Ω r = ρ 1 g ( η r E ) + ρ g E r r ρ Ω r = ρ 1 g η E ρg r r ρ Ω r = ρ 1 gω r ρg E r ρg E = ρω r r E r = Ω g r Thus, the interface height is increasing away from the center. It will have the same shape as the free surface. This is consistent with the idea that geopotential surfaces depend only on gravity and centrifugal force, and therefore are not dependent on density. 5
6 .a. First consider the pendulum in a non-rotating reference frame. The acceleration a of the pendulum displaced an angle θ from its resting position is ( x a = g sin θ = g l i + y ) l j Since the length of the string is long, we can assume almost entirely horizontal accelerations such that a = du i + dv j: du dv ( g ) = l x = ω x ( g ) = l y = ω y Defining the downward direction as the direction of effective gravity, the Coriolis force will be the only apparent force in the rotating frame. Neglecting small vertical velocities, this force is Ω u = fvi fuj. 6
7 The full momentum equation in the rotating reference frame (with no centrifugal acceleration term) will therefore be du r = du f Ω u. Substituting the acceleration in the fixed frame and the Coriolis force gives horizontal components du dv = ω x + fv = ω y fu It is easiest to solve these equations by defining a complex position s = x + iy and obtaining one ordinary differential equation: d s = du + idv = ( ω x + fv) + i( ω y fu) = ω (x + iy) if(u + iv) = ω s if ds This equation can be solved by using the method of characteristics to find eigenvalues λ 1 and λ : d s ds + if + ω s = 0 λ + ifλ + ω = 0 Using the quadratic equation, gives λ 1 = if i f +4ω and λ = if + f +4ω i. Let us define ω = then write the solution as: f +4ω. Note that if f << ω ω ω. We can s = Ae λ1t + Be λt = Ae ( if iω )t + BAe ( if +iω )t = (Ae iω t + Be iω t )e i f t 7
8 where A and B are constants. These constants can be determined by plugging in the initial conditions. At t = 0, s = a, so a = A + B. At time t = 0 there is no initial velocity, ds = 0 = ( if iω )A + ( if + iω )B A = B (ω f/) (ω + f/) a B = B (ω f/) (ω + f/) [ (ω ] f/) a = B (ω + f/) + 1 [ ω ] a = B ω + f/ B = a [ ω ] + f/ ω [ ω ] f/ A = a ω With these constants determined, we see that A + B = a and B A = f/ ω. Now, rearranging the equation for s: s = (Ae iω t + Be iω t )e i f t = [A cos (ω t) ia sin (ω t) + B cos (ω t) + ib sin (ω t)] (cos ( f ) t) i sin (f t) = (A + B) cos (ω t) (cos ( f ) t) i sin (f t) + f/ ω (B A) sin (ω t) (sin ( f ) t) + i cos (f t) = a cos (ω t) (cos ( f ) t) i sin (f t) + f/ ω sin (ω t) (sin ( f ) t) + i cos (f t) Of the two terms on the RHS, the first term shows that clockwise circular motion (in the northern hemisphere) at the frequency f/ = Ω sin φ will be superimposed on the back and forth motion at motions at frequency ω, signified by a cos ωt. The second term modifies the path of the pendulum only slightly in the case a pendulum on the earth, where f/ ω should be very small for a realistically scaled pendulum. In order to determine latitude from a Foucault pendulum, simply observe how long it takes for the precession of the pendulum to complete a full circle. The time it takes for the pendulum to complete this full circle is equal to 4π f. The latitude is equal to 8
9 ( ) ( ) f 86400s φ = sin 1 = sin 1. Ω 4π/f Here Ω is the earth s rotation frequency and φ is latitude. Note: the pendulum will swing back toward a fixed observer at one half the full inertial period π/f, when the pendulum has completed one half of a full circle. 9
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